Let show that every has a removable singularity at .
I'm quite certain this is an easy question, but I'm lacking the background for the appropriate proof. Any help?
Since no one apparently knows this, let me go ahead and type this:
z=0 is a removable singularity of if there exists a holomorphic function g definied on such that . Hence let . f is holomorphic on , since f is holomorphic on G.
Am I missing something? It seems that by definition of being holomorphic on G, you're holomorphic at {0}, no?
This is a nice problem, unfortunately I don't have a full solution yet.
It's easy to see that f can't have a pole (because then with and iff ), but I haven't been able to deduce the case where f has an essential singularity (It's informally clear since f takes on (almost) the whole plane an infinite number of times in any neighbourhood of 0, but the proof eludes me)